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Math Help - Polynomial Rings - Z5[X]/<p(x)>

  1. #1
    Super Member Bernhard's Avatar
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    Polynomial Rings - Z5[X]/<p(x)>

    Compute the multiplicative inverse of  a(x) = \overline{x^2 + x + 1} in  \mathbb{Z}_5/<p(x)>

    where  p(x) = x^3 + x^2 + x + 1 = (x^2 + 1)(x +1) .
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  2. #2
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    Re: Polynomial Rings - Z5[X]/<p(x)>

    Quote Originally Posted by Bernhard View Post
    Compute the multiplicative inverse of  a(x) = \overline{x^2 + x + 1} in  \mathbb{Z}_5/<p(x)>

    where  p(x) = x^3 + x^2 + x + 1 = (x^2 + 1)(x +1) .
    Because x^3+x^2+x+1=0, we have 1=-x^3-x^2-x. It is easy to see that (-x)(x^2+x+1)=-x^3-x^2-x, so -x is the inverse.



    However, it is not always so easy to solve for inverses by inspection. The surefire way to do this is to use linear algebra, as I will demonstrate.

    You want (a+bx+cx^2)(x^2+x+1)=1 (we only go up to the second degree since x^3=-x^2-x-1)

    Expanding the left hand side gives

    \bigg(ax^2+ax+a\bigg)+\bigg(bx^3+bx^2+bx\bigg)+ \bigg(cx^4+cx^3+cx^2\bigg)

    =\bigg(a(x^3+x^2+x+1) - ax^3\bigg)
    +\bigg(b(x^3+x^2+x+1)-b\bigg)
    + \bigg(cx(x^3+x^2+x+1)-cx\bigg)

    =-ax^3-cx-b

    =a(x^2+x+1)-cx-b

    =ax^2+(a-c)x+(a-b)



    So we want to find a,b,c\in \mathbb{Z}_5 such that ax^2+(a-c)x+(a-b)=1. Although it is obvious what the solution is in this case, I'll write down the general way to do it:

    With respect to the basis {1,x,x^2}, this is equivalent to solving the linear system

    \begin{bmatrix} 1&-1&0 \\ 1&0&-1 \\1&0&0 \end{bmatrix}\begin{bmatrix} a  \\ b \\c \end{bmatrix}=\begin{bmatrix} 1  \\ 0 \\0 \end{bmatrix}

    Which has solutions a=c=0;b=-1. Hence the inverse is ax^2+bx+c=-x, which is the same answer as we got from inspection.
    Thanks from Bernhard
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  3. #3
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    Re: Polynomial Rings - Z5[X]/<p(x)>

    How do you guys post equations so neatly?
    By copying and pasting the original post, this is what I get:

    "Compute the multiplicative inverse of a(x) = \overline{x^2 + x + 1} in \mathbb{Z}_5/<p(x)>

    where p(x) = x^3 + x^2 + x + 1 = (x^2 + 1)(x +1) . "
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  4. #4
    Super Member Bernhard's Avatar
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    Re: Polynomial Rings - Z5[X]/<p(x)>

    Hi HowDoIMath,

    You need to use the Latex language for Maths expressions - see Latex Help and download the very helpful tutorials or help sheets.

    Note that you need [TEX] and then the same with a / before the T around the math expressions to get Latex to work

    Peter
    Last edited by Bernhard; April 6th 2013 at 11:40 PM.
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  5. #5
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    Re: Polynomial Rings - Z5[X]/<p(x)>

    Ok thanks
    a(x) = \overline{x^2 + x + 1}
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